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| Question 1180470:  One solution to the system of equations :  y= mx^2 +nx +n and y= nx^2 -mx +n is (-2,11) . Algebraically determine the values of m and n.
 I appreciate any help. Thank you:)
 Found 2 solutions by  ikleyn, MathLover1:
 Answer by ikleyn(52878)
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You can put this solution on YOUR website! . One solution to the system of equations : y= mx^2 +nx +n and y= nx^2 -mx +n is (-2,11) . Algebraically determine the values of m and n.
 I appreciate any help. Thank you:)
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First, substitute the solution x= -2, y= 11 into the first equation.  You will get then
    11 = m*(-2)^2 + n*(-2) + n,
or, equivalently,
    11 = 4m - n.    (1)
Next, substitute the solution x= -2, y= 11 into the second equation.  You will get then
    11 = n*(-2)^2 - m*(-2) + n,
or, equivalently,
    11 = 5n + 2m.    (2)
Equations (1) and (2) form the system of 2 equations in 2 unknowns m and n to solve. 
I will re-write the system in the standard form
    4m -  n = 11      (1')
    2m + 5n = 11      (2')
To solve the system, from equation (1') express  n = 4m-11  and substitute it into (2').  You will get
    2m + 5*(4m-11) = 11
Simplify and solve
    2m + 20m - 55 = 11
    22m            = 11 + 55 = 66
      m                      = 66/22 = 3.
Finally, from n = 4m-11 you find  n = 4*3-11 = 12-11 = 1.
ANSWER.  n = 1;  m = 3.
Solved.
 
 You may CHECK it on your own, that my solution is correct, by substituting the found values
 into the original equations.
 
 
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Answer by MathLover1(20850)
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